A Classical Approach of Newtonian Mechanics

8 ROTATIONAL MOTION 8.11 Combined translational and rotational motion

M g sin θ

f =

1 + M b^2 /I

. (8.100)

Since the moment of inertia of the cylinder is actually I = (1/2) M b^2 , the above
expressions simplify to give

and

̇v =

2

g sin θ, (8.101)

3

f =

1
M g sin θ. (8.102)
3
Note that the acceleration of a uniform cylinder as it rolls down a slope, without

slipping, is only two-thirds of the value obtained when the cylinder slides down
the same slope without friction. It is clear from Eq. (8.95) that, in the former

case, the acceleration of the cylinder down the slope is retarded by friction. Note,

however, that the frictional force merely acts to convert translational kinetic en-

ergy into rotational kinetic energy, and does not dissipate energy.

Now, in order for the slope to exert the frictional force specified in Eq. (8.102),
without any slippage between the slope and cylinder, this force must be less than

the maximum allowable static frictional force, μ R(= μ M g cos θ), where μ is the
coefficient of static friction. In other words, the condition for the cylinder to roll

down the slope without slipping is f < μ R, or

tan θ < 3 μ. (8.103)

This condition is easily satisfied for gentle slopes, but may well be violated for ex-

tremely steep slopes (depending on the size of μ). Of course, the above condition

is always violated for frictionless slopes, for which μ = 0.

Suppose, finally, that we place two cylinders, side by side and at rest, at the top

of a frictional slope of inclination θ. Let the two cylinders possess the same mass,

M, and the same radius, b. However, suppose that the first cylinder is uniform,

whereas the second is a hollow shell. Which cylinder reaches the bottom of

the slope first, assuming that they are both released simultaneously, and both

roll without slipping? The acceleration of each cylinder down the slope is given
by Eq. (8.99). For the case of the solid cylinder, the moment of inertia is I =